Solve : $4x +\dfrac{6}{y} = 15$ and $6x - \dfrac{8}{y} = 14$.

Hence, find the value of 'k', if y = kx - 2.

Lets take $\dfrac{1}{y}$ = u, we get

4x + 6u = 15

6x - 8u = 14

Multiply first equation by 3 and second equation by 2, then subtract both equations.

(4x + 6u = 15) x 3

(6x - 8u = 14) x 2

$\begin{matrix} & 12x & + & 18u & = & 45 \ & 12u & - & 16u & = & 28 \ & - & + & & & - \ \hline & & & 34u & = & 45 - 28 \ \Rightarrow & & & 34u & = & 17 \end{matrix}$

⇒ u = $\dfrac{17}{34}$

⇒ u = $\dfrac{1}{2}$

Substituting the value of u in first equation, we get:

⇒ 4x + 6 $\times \dfrac{1}{2}$ = 15

⇒ 4x + 3 = 15

⇒ 4x = 15 - 3

⇒ 4x = 12

⇒ x = $\dfrac{12}{4}$

⇒ x = 3

So, y = $\dfrac{1}{u}$ = 2

Now, put the value of x and y in y = kx - 2,

⇒ 2 = k $\times$ 3 - 2

⇒ 2 + 2 = k $\times$ 3

⇒ k $\times$ 3 = 4

⇒ k = $\dfrac{4}{3}$

Hence, the value of x = 3 , y = 2 and k = $\dfrac{4}{3}$.